A ball is projected with kinetic energy E, at an angle of 60° to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Solution

Physics sir · 3 steps

At the top only the horizontal velocity survives, so the KE drops by a factor of cos² 60°.

1. Velocity at the top. Gravity acts only vertically, so the horizontal component u cos 60° never changes. At the highest point the vertical component is zero, so the speed there is just u cos 60° = u/2.

   E' = 1/2m(u cos 60°)² = (1/2mu²)cos² 60° = E cos² 60°

2. Put in numbers. cos 60° = 1/2, so cos² 60° = 1/4. Kinetic energy depends on the square of speed, so halving the speed quarters the energy.
3. So. The kinetic energy at the highest point is E/4, option (c).

The idea to keep. At the top of a projectile's path, KE = E cos² θ. The ball never stops: its horizontal velocity carries straight through the whole flight.

Answer from the NTA final answer key. Solution by PrepWiser.